biased samples (in honor of prof. c.r. rao)...biased samples (in honor of prof. c.r. rao) jorge...

Biased samplesRenewal processesHow to detect biased samples?Appendix

Biased samples (in honor of Prof. C.R. Rao)Jorge Navarro1;2

1Universidad de Murcia, Spain. E-mail: [email protected] is partially supported by Ministerio de Ciencia y Tecnolog��a de Espa~naunder grant MTM2009-08311 and Fundaci�on S�eneca (C.A.R.M.) under grant08627/PI/08. Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?AppendixBiased samplesDe�nitionRao's exampleFisher's exampleRenewal processesWaiting time paradoxEquilibrium distributionHow to detect biased samples?Mean sojourn time per touristHow to be a rich man?AppendixConclusionsSome of my referencesOther references

Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleBiased and censored samples

I X1; :::;Xn sample from X

I X1; :::;Xn i.i.d. Pr(Xi � x) = Pr(X � x)I Censored sample: Some Xi are unknown.I Example: X =lifetime of...2; 3; 5; 6; 7; :::; 1+; 3+; 4+; :::

I 1+ means Xi > 1I Biased sample: the sample probability of Xi depends on Xi .I Example: A sample from families recover from their children.I Censored samples are a particular case.



I X1; :::;Xn sample from XI X1; :::;Xn i.i.d. Pr(Xi � x) = Pr(X � x)

I Censored sample: Some Xi are unknown.I Example: X =lifetime of...2; 3; 5; 6; 7; :::; 1+; 3+; 4+; :::




I X1; :::;Xn sample from XI X1; :::;Xn i.i.d. Pr(Xi � x) = Pr(X � x)I Censored sample: Some Xi are unknown.

I Example: X =lifetime of...2; 3; 5; 6; 7; :::; 1+; 3+; 4+; :::




I X1; :::;Xn sample from XI X1; :::;Xn i.i.d. Pr(Xi � x) = Pr(X � x)I Censored sample: Some Xi are unknown.I Example: X =lifetime of...2; 3; 5; 6; 7; :::; 1+; 3+; 4+; :::





I 1+ means Xi > 1

I Biased sample: the sample probability of Xi depends on Xi .I Example: A sample from families recover from their children.I Censored samples are a particular case.




I 1+ means Xi > 1I Biased sample: the sample probability of Xi depends on Xi .

I Example: A sample from families recover from their children.I Censored samples are a particular case.




I 1+ means Xi > 1I Biased sample: the sample probability of Xi depends on Xi .I Example: A sample from families recover from their children.

I Censored samples are a particular case.




I 1+ means Xi > 1I Biased sample: the sample probability of Xi depends on Xi .I Example: A sample from families recover from their children.I Censored samples are a particular case.Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleA model for biased samples

I First example: Fisher (1934, Ann. Eugenics 6, 13-25).

I Model: C.R. Rao (1965, Sankhya Ser. A 27, 311-324).I Y has the biased (or weighted) distribution associated to Xand w(t) � 0 iffY (t) = w(t)fX (t)E (w(X ))I With this model the probability of observe Xi = t isproportional to w(t).I How to study X from a sample Y1; :::;Yn from Y ?I Censored data in A: w(t) = 1 if t 2 A (0 elsewhere).I Biased data: the probability of observe Xi is proportional tow(Xi ).



I First example: Fisher (1934, Ann. Eugenics 6, 13-25).I Model: C.R. Rao (1965, Sankhya Ser. A 27, 311-324).

I Y has the biased (or weighted) distribution associated to Xand w(t) � 0 iffY (t) = w(t)fX (t)E (w(X ))I With this model the probability of observe Xi = t isproportional to w(t).I How to study X from a sample Y1; :::;Yn from Y ?I Censored data in A: w(t) = 1 if t 2 A (0 elsewhere).I Biased data: the probability of observe Xi is proportional tow(Xi ).



I First example: Fisher (1934, Ann. Eugenics 6, 13-25).I Model: C.R. Rao (1965, Sankhya Ser. A 27, 311-324).I Y has the biased (or weighted) distribution associated to Xand w(t) � 0 iffY (t) = w(t)fX (t)E (w(X ))

I With this model the probability of observe Xi = t isproportional to w(t).I How to study X from a sample Y1; :::;Yn from Y ?I Censored data in A: w(t) = 1 if t 2 A (0 elsewhere).I Biased data: the probability of observe Xi is proportional tow(Xi ).



I First example: Fisher (1934, Ann. Eugenics 6, 13-25).I Model: C.R. Rao (1965, Sankhya Ser. A 27, 311-324).I Y has the biased (or weighted) distribution associated to Xand w(t) � 0 iffY (t) = w(t)fX (t)E (w(X ))I With this model the probability of observe Xi = t isproportional to w(t).

I How to study X from a sample Y1; :::;Yn from Y ?I Censored data in A: w(t) = 1 if t 2 A (0 elsewhere).I Biased data: the probability of observe Xi is proportional tow(Xi ).



I First example: Fisher (1934, Ann. Eugenics 6, 13-25).I Model: C.R. Rao (1965, Sankhya Ser. A 27, 311-324).I Y has the biased (or weighted) distribution associated to Xand w(t) � 0 iffY (t) = w(t)fX (t)E (w(X ))I With this model the probability of observe Xi = t isproportional to w(t).I How to study X from a sample Y1; :::;Yn from Y ?

I Censored data in A: w(t) = 1 if t 2 A (0 elsewhere).I Biased data: the probability of observe Xi is proportional tow(Xi ).



I First example: Fisher (1934, Ann. Eugenics 6, 13-25).I Model: C.R. Rao (1965, Sankhya Ser. A 27, 311-324).I Y has the biased (or weighted) distribution associated to Xand w(t) � 0 iffY (t) = w(t)fX (t)E (w(X ))I With this model the probability of observe Xi = t isproportional to w(t).I How to study X from a sample Y1; :::;Yn from Y ?I Censored data in A: w(t) = 1 if t 2 A (0 elsewhere).

I Biased data: the probability of observe Xi is proportional tow(Xi ).



I First example: Fisher (1934, Ann. Eugenics 6, 13-25).I Model: C.R. Rao (1965, Sankhya Ser. A 27, 311-324).I Y has the biased (or weighted) distribution associated to Xand w(t) � 0 iffY (t) = w(t)fX (t)E (w(X ))I With this model the probability of observe Xi = t isproportional to w(t).I How to study X from a sample Y1; :::;Yn from Y ?I Censored data in A: w(t) = 1 if t 2 A (0 elsewhere).I Biased data: the probability of observe Xi is proportional tow(Xi ).


Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleRao's example

I Rao (1977, American Statistician 31, 24-26).

I In a survey we ask for the number of brother and sisters((�)including yourself):Sex Brothers� Sisters� TotalM or W Yi Xi mi = Xi + Yi- - - -

I Predictions (sample from men)

1. M =PYi >>W =PXi2. M �W =PYi �PXi ' k = n (sample size)3. M=N = (PYi )=(Pm i ) >> 0:5; N =Pmi = M +W4. M=N = (PYi )=(Pm i ) ' 0:5 + k2Pmi5. M � kN � k = PYi � kPmi � k ' 0:5



I Rao (1977, American Statistician 31, 24-26).I In a survey we ask for the number of brother and sisters((�)including yourself):Sex Brothers� Sisters� TotalM or W Yi Xi mi = Xi + Yi- - - -

I Predictions (sample from men)

1. M =PYi >>W =PXi2. M �W =PYi �PXi ' k = n (sample size)3. M=N = (PYi )=(Pm i ) >> 0:5; N =Pmi = M +W4. M=N = (PYi )=(Pm i ) ' 0:5 + k2Pmi5. M � kN � k = PYi � kPmi � k ' 0:5




I Predictions (sample from men)1. M =PYi >>W =PXi

2. M �W =PYi �PXi ' k = n (sample size)3. M=N = (PYi )=(Pm i ) >> 0:5; N =Pmi = M +W4. M=N = (PYi )=(Pm i ) ' 0:5 + k2Pmi5. M � kN � k = PYi � kPmi � k ' 0:5




I Predictions (sample from men)1. M =PYi >>W =PXi2. M �W =PYi �PXi ' k = n (sample size)

3. M=N = (PYi )=(Pm i ) >> 0:5; N =Pmi = M +W4. M=N = (PYi )=(Pm i ) ' 0:5 + k2Pmi5. M � kN � k = PYi � kPmi � k ' 0:5




I Predictions (sample from men)1. M =PYi >>W =PXi2. M �W =PYi �PXi ' k = n (sample size)3. M=N = (PYi )=(Pm i ) >> 0:5; N =Pmi = M +W

4. M=N = (PYi )=(Pm i ) ' 0:5 + k2Pmi5. M � kN � k = PYi � kPmi � k ' 0:5




I Predictions (sample from men)1. M =PYi >>W =PXi2. M �W =PYi �PXi ' k = n (sample size)3. M=N = (PYi )=(Pm i ) >> 0:5; N =Pmi = M +W4. M=N = (PYi )=(Pm i ) ' 0:5 + k2Pmi

5. M � kN � k = PYi � kPmi � k ' 0:5




I Predictions (sample from men)1. M =PYi >>W =PXi2. M �W =PYi �PXi ' k = n (sample size)3. M=N = (PYi )=(Pm i ) >> 0:5; N =Pmi = M +W4. M=N = (PYi )=(Pm i ) ' 0:5 + k2Pmi5. M � kN � k = PYi � kPmi � k ' 0:5


Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleRao's results

City N M W M-W k M/N 12 + k2N M�kN�kTehran 105 65 40 25 21 0.619 0.600 0.524Isphahan 77 45 32 13 11 0.584 0.571 0.515Tokyo 124 90 34 56 50 0.726 0.701 0.540Delhi 158 92 66 26 29 0.582 0.592 0.488Calcutta 726 414 312 102 104 0.570 0.571 0.498Waltair 211 123 88 35 39 0.583 0.592 0.488Ahmed. 133 84 49 35 29 0.632 0.609 0.529Bangalore 307 180 127 53 55 0.586 0.589 0.496


Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleQuestions

I How to estimate pH or pM?

I How to estimate E (mi )?I Which sample is the best one?I Can we use both samples together?I How can we obtain the best results?



I How to estimate pH or pM?I How to estimate E (mi )?

I Which sample is the best one?I Can we use both samples together?I How can we obtain the best results?



I How to estimate pH or pM?I How to estimate E (mi )?I Which sample is the best one?

I Can we use both samples together?I How can we obtain the best results?



I How to estimate pH or pM?I How to estimate E (mi )?I Which sample is the best one?I Can we use both samples together?

I How can we obtain the best results?



I How to estimate pH or pM?I How to estimate E (mi )?I Which sample is the best one?I Can we use both samples together?I How can we obtain the best results?


Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleSolutions

I The number of brothers is a Binomial B(m; pM), withpM ' 0:5p(x) = Pr(X = x) = �mx

�pxM � pm�xWE (X ) = mpM

I The sampling probability of Yi is proportional to Yi .I Hence Y is a length biased Binomial Y � B�(mi ; pM)p�(x) = xp(x)E (X ) = x�mix

�pxM � pmi�xW =(mipM)= x xmi !mix!(mi � x)!px�1M pmi�xW = �mi � 1x � 1

�px�1M pm�xW ; x = 1; 2; :::I Yi � 1 � B(mi � 1; pM)




�pxM � pm�xWE (X ) = mpMI The sampling probability of Yi is proportional to Yi .

I Hence Y is a length biased Binomial Y � B�(mi ; pM)p�(x) = xp(x)E (X ) = x�mix

�pxM � pmi�xW =(mipM)= x xmi !mix!(mi � x)!px�1M pmi�xW = �mi � 1x � 1

�px�1M pm�xW ; x = 1; 2; :::I Yi � 1 � B(mi � 1; pM)




�pxM � pm�xWE (X ) = mpMI The sampling probability of Yi is proportional to Yi .I Hence Y is a length biased Binomial Y � B�(mi ; pM)

p�(x) = xp(x)E (X ) = x�mix�pxM � pmi�xW =(mipM)

= x xmi !mix!(mi � x)!px�1M pmi�xW = �mi � 1x � 1�px�1M pm�xW ; x = 1; 2; :::

I Yi � 1 � B(mi � 1; pM)




�pxM � pm�xWE (X ) = mpMI The sampling probability of Yi is proportional to Yi .I Hence Y is a length biased Binomial Y � B�(mi ; pM)

p�(x) = xp(x)E (X ) = x�mix�pxM � pmi�xW =(mipM)

= x xmi !mix!(mi � x)!px�1M pmi�xW = �mi � 1x � 1�px�1M pm�xW ; x = 1; 2; :::

I Yi � 1 � B(mi � 1; pM)Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's examplePredictions:

I Yi � 1 � B(mi � 1; pM)

I E (Yi ) = 1 + (mi � 1)pM = 1� pM +mipMI Xi � B(mi � 1; pW )I E (Xi ) = (mi � 1)pWI E (PYi ) =PE (Yi ) =P(1� pW +mipM) =k(1� pM) + pMPmiI E (PXi ) =PE (Xi ) =P(mi � 1)pW = pW Pmi � kpWI E (PYi �PXi ) = 2kpW ' kI E �PYiPmi

� = kpW + pMPmiPmi = pM+ kpWPmi ' 0:5+ k2PmiI E �PYi � kPmi � k

� = k(1� pM) + pMPmi � kPmi � k = pM ' 0:5



I Yi � 1 � B(mi � 1; pM)I E (Yi ) = 1 + (mi � 1)pM = 1� pM +mipM

I Xi � B(mi � 1; pW )I E (Xi ) = (mi � 1)pWI E (PYi ) =PE (Yi ) =P(1� pW +mipM) =k(1� pM) + pMPmiI E (PXi ) =PE (Xi ) =P(mi � 1)pW = pW Pmi � kpWI E (PYi �PXi ) = 2kpW ' kI E �PYiPmi





I Yi � 1 � B(mi � 1; pM)I E (Yi ) = 1 + (mi � 1)pM = 1� pM +mipMI Xi � B(mi � 1; pW )

I E (Xi ) = (mi � 1)pWI E (PYi ) =PE (Yi ) =P(1� pW +mipM) =k(1� pM) + pMPmiI E (PXi ) =PE (Xi ) =P(mi � 1)pW = pW Pmi � kpWI E (PYi �PXi ) = 2kpW ' kI E �PYiPmi





I Yi � 1 � B(mi � 1; pM)I E (Yi ) = 1 + (mi � 1)pM = 1� pM +mipMI Xi � B(mi � 1; pW )I E (Xi ) = (mi � 1)pW

I E (PYi ) =PE (Yi ) =P(1� pW +mipM) =k(1� pM) + pMPmiI E (PXi ) =PE (Xi ) =P(mi � 1)pW = pW Pmi � kpWI E (PYi �PXi ) = 2kpW ' kI E �PYiPmi





I Yi � 1 � B(mi � 1; pM)I E (Yi ) = 1 + (mi � 1)pM = 1� pM +mipMI Xi � B(mi � 1; pW )I E (Xi ) = (mi � 1)pWI E (PYi ) =PE (Yi ) =P(1� pW +mipM) =k(1� pM) + pMPmi

I E (PXi ) =PE (Xi ) =P(mi � 1)pW = pW Pmi � kpWI E (PYi �PXi ) = 2kpW ' kI E �PYiPmi





I Yi � 1 � B(mi � 1; pM)I E (Yi ) = 1 + (mi � 1)pM = 1� pM +mipMI Xi � B(mi � 1; pW )I E (Xi ) = (mi � 1)pWI E (PYi ) =PE (Yi ) =P(1� pW +mipM) =k(1� pM) + pMPmiI E (PXi ) =PE (Xi ) =P(mi � 1)pW = pW Pmi � kpW

I E (PYi �PXi ) = 2kpW ' kI E �PYiPmi





I Yi � 1 � B(mi � 1; pM)I E (Yi ) = 1 + (mi � 1)pM = 1� pM +mipMI Xi � B(mi � 1; pW )I E (Xi ) = (mi � 1)pWI E (PYi ) =PE (Yi ) =P(1� pW +mipM) =k(1� pM) + pMPmiI E (PXi ) =PE (Xi ) =P(mi � 1)pW = pW Pmi � kpWI E (PYi �PXi ) = 2kpW ' k

I E �PYiPmi� = kpW + pMPmiPmi = pM+ kpWPmi ' 0:5+ k2Pmi

I E �PYi � kPmi � k� = k(1� pM) + pMPmi � kPmi � k = pM ' 0:5



I Yi � 1 � B(mi � 1; pM)I E (Yi ) = 1 + (mi � 1)pM = 1� pM +mipMI Xi � B(mi � 1; pW )I E (Xi ) = (mi � 1)pWI E (PYi ) =PE (Yi ) =P(1� pW +mipM) =k(1� pM) + pMPmiI E (PXi ) =PE (Xi ) =P(mi � 1)pW = pW Pmi � kpWI E (PYi �PXi ) = 2kpW ' kI E �PYiPmi

� = kpW + pMPmiPmi = pM+ kpWPmi ' 0:5+ k2Pmi

I E �PYi � kPmi � k� = k(1� pM) + pMPmi � kPmi � k = pM ' 0:5



I Yi � 1 � B(mi � 1; pM)I E (Yi ) = 1 + (mi � 1)pM = 1� pM +mipMI Xi � B(mi � 1; pW )I E (Xi ) = (mi � 1)pWI E (PYi ) =PE (Yi ) =P(1� pW +mipM) =k(1� pM) + pMPmiI E (PXi ) =PE (Xi ) =P(mi � 1)pW = pW Pmi � kpWI E (PYi �PXi ) = 2kpW ' kI E �PYiPmi


� = k(1� pM) + pMPmi � kPmi � k = pM ' 0:5Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)


I How to estimate pM?

I We can use:T = PYi � kPmi � k

E (T ) = E (PYi � kPmi � k ) = pMVat(T ) = pMpW =(Xmi � k)! 0XYi � k � B(Xmi � k ; pM)

T u NormalT is an UMVUE



I How to estimate pM?I We can use:T = PYi � kPmi � k

E (T ) = E (PYi � kPmi � k ) = pMVat(T ) = pMpW =(Xmi � k)! 0XYi � k � B(Xmi � k ; pM)

T u NormalT is an UMVUEJorge Navarro; Biased samples (in honor of Prof. C.R. Rao)


I How to use both samples?

I Let X1; :::;Xn be an unbiased sample from Xi � B(ni ; p).I Let Y1; :::;Ym be a length biased sample.I Then Yj � 1 � B(mj � 1; p) andT = PXi +P(Yj � 1)P ni +P(mj � 1)

E (T ) = E (PXi +P(Yj � 1)P ni +P(mj � 1)) = pVat(T ) = p(1� p)=(X ni �X(mj � 1))XXi +X(Yj � 1) � B(X ni �X(mj � 1); pM)

T u NormalT is UMVUE



I How to use both samples?I Let X1; :::;Xn be an unbiased sample from Xi � B(ni ; p).

I Let Y1; :::;Ym be a length biased sample.I Then Yj � 1 � B(mj � 1; p) andT = PXi +P(Yj � 1)P ni +P(mj � 1)





I How to use both samples?I Let X1; :::;Xn be an unbiased sample from Xi � B(ni ; p).I Let Y1; :::;Ym be a length biased sample.

I Then Yj � 1 � B(mj � 1; p) andT = PXi +P(Yj � 1)P ni +P(mj � 1)





I How to use both samples?I Let X1; :::;Xn be an unbiased sample from Xi � B(ni ; p).I Let Y1; :::;Ym be a length biased sample.I Then Yj � 1 � B(mj � 1; p) andT = PXi +P(Yj � 1)P ni +P(mj � 1)


T u NormalT is UMVUEJorge Navarro; Biased samples (in honor of Prof. C.R. Rao)


I What is the best sample?

I If Yj = 1; then the information is null.I Xi has more information than Yj if ni > mj � 1I The Fisher's information (I1 = E [( @@pp(x))2]) are:IXi (p) = nipqIYj (p) = mj � 1pq

I E (ni ) =?;E (mj) =? (mj � 1)I In our survey mj � 1 = nj , so both samples have the sameinformation (in each data).I The best option is to use both samples together!



I What is the best sample?I If Yj = 1; then the information is null.

I Xi has more information than Yj if ni > mj � 1I The Fisher's information (I1 = E [( @@pp(x))2]) are:IXi (p) = nipqIYj (p) = mj � 1pq




I What is the best sample?I If Yj = 1; then the information is null.I Xi has more information than Yj if ni > mj � 1

I The Fisher's information (I1 = E [( @@pp(x))2]) are:IXi (p) = nipqIYj (p) = mj � 1pq




I What is the best sample?I If Yj = 1; then the information is null.I Xi has more information than Yj if ni > mj � 1I The Fisher's information (I1 = E [( @@pp(x))2]) are:IXi (p) = nipqIYj (p) = mj � 1pq





I E (ni ) =?;E (mj) =? (mj � 1)

I In our survey mj � 1 = nj , so both samples have the sameinformation (in each data).I The best option is to use both samples together!




I E (ni ) =?;E (mj) =? (mj � 1)I In our survey mj � 1 = nj , so both samples have the sameinformation (in each data).

I The best option is to use both samples together!




I E (ni ) =?;E (mj) =? (mj � 1)I In our survey mj � 1 = nj , so both samples have the sameinformation (in each data).I The best option is to use both samples together!Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleAdditional questions

I How to estimate the number of children m?

I Can we use m = 1k Pmi?I If we use men and women, the sampling probability of afamily with mi children is proportional to miI If we only use men, it is proportional to E (Xi ) = mipMI Then m1; :::;mk is a length biased sample from m.I How to estimate E (m) using m1; :::;mk?



I How to estimate the number of children m?I Can we use m = 1k Pmi?

I If we use men and women, the sampling probability of afamily with mi children is proportional to miI If we only use men, it is proportional to E (Xi ) = mipMI Then m1; :::;mk is a length biased sample from m.I How to estimate E (m) using m1; :::;mk?



I How to estimate the number of children m?I Can we use m = 1k Pmi?I If we use men and women, the sampling probability of afamily with mi children is proportional to mi

I If we only use men, it is proportional to E (Xi ) = mipMI Then m1; :::;mk is a length biased sample from m.I How to estimate E (m) using m1; :::;mk?



I How to estimate the number of children m?I Can we use m = 1k Pmi?I If we use men and women, the sampling probability of afamily with mi children is proportional to miI If we only use men, it is proportional to E (Xi ) = mipM

I Then m1; :::;mk is a length biased sample from m.I How to estimate E (m) using m1; :::;mk?



I How to estimate the number of children m?I Can we use m = 1k Pmi?I If we use men and women, the sampling probability of afamily with mi children is proportional to miI If we only use men, it is proportional to E (Xi ) = mipMI Then m1; :::;mk is a length biased sample from m.

I How to estimate E (m) using m1; :::;mk?



I How to estimate the number of children m?I Can we use m = 1k Pmi?I If we use men and women, the sampling probability of afamily with mi children is proportional to miI If we only use men, it is proportional to E (Xi ) = mipMI Then m1; :::;mk is a length biased sample from m.I How to estimate E (m) using m1; :::;mk?



I If m � Poisson(�); � =mean number of childrenp(x) = �xe��=x!; x = 0; 1; :::

I Hence mj � size biased Poisson withp�(x) = xp(x)� = x�xe��

�x! = �x�1e��((x � 1)! ; x = 1; 2; :::

I Then mj � 1 � Poisson(�)I E (m) = 1k PE (mi ) = 1k P(�+ 1) = �+ 1I T = m � 1 = 1k P(mi � 1)I E (T ) = �I Var(T ) = �=k ! 0I P(mi � 1) � Poisson(k�)I T u Normal





�x! = �x�1e��((x � 1)! ; x = 1; 2; :::

I Then mj � 1 � Poisson(�)

I E (m) = 1k PE (mi ) = 1k P(�+ 1) = �+ 1I T = m � 1 = 1k P(mi � 1)I E (T ) = �I Var(T ) = �=k ! 0I P(mi � 1) � Poisson(k�)I T u Normal





�x! = �x�1e��((x � 1)! ; x = 1; 2; :::

I Then mj � 1 � Poisson(�)I E (m) = 1k PE (mi ) = 1k P(�+ 1) = �+ 1

I T = m � 1 = 1k P(mi � 1)I E (T ) = �I Var(T ) = �=k ! 0I P(mi � 1) � Poisson(k�)I T u Normal





�x! = �x�1e��((x � 1)! ; x = 1; 2; :::

I Then mj � 1 � Poisson(�)I E (m) = 1k PE (mi ) = 1k P(�+ 1) = �+ 1I T = m � 1 = 1k P(mi � 1)

I E (T ) = �I Var(T ) = �=k ! 0I P(mi � 1) � Poisson(k�)I T u Normal





�x! = �x�1e��((x � 1)! ; x = 1; 2; :::

I Then mj � 1 � Poisson(�)I E (m) = 1k PE (mi ) = 1k P(�+ 1) = �+ 1I T = m � 1 = 1k P(mi � 1)I E (T ) = �

I Var(T ) = �=k ! 0I P(mi � 1) � Poisson(k�)I T u Normal





�x! = �x�1e��((x � 1)! ; x = 1; 2; :::

I Then mj � 1 � Poisson(�)I E (m) = 1k PE (mi ) = 1k P(�+ 1) = �+ 1I T = m � 1 = 1k P(mi � 1)I E (T ) = �I Var(T ) = �=k ! 0

I P(mi � 1) � Poisson(k�)I T u Normal





�x! = �x�1e��((x � 1)! ; x = 1; 2; :::

I Then mj � 1 � Poisson(�)I E (m) = 1k PE (mi ) = 1k P(�+ 1) = �+ 1I T = m � 1 = 1k P(mi � 1)I E (T ) = �I Var(T ) = �=k ! 0I P(mi � 1) � Poisson(k�)

I T u Normal





�x! = �x�1e��((x � 1)! ; x = 1; 2; :::

I Then mj � 1 � Poisson(�)I E (m) = 1k PE (mi ) = 1k P(�+ 1) = �+ 1I T = m � 1 = 1k P(mi � 1)I E (T ) = �I Var(T ) = �=k ! 0I P(mi � 1) � Poisson(k�)I T u Normal Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)


I Results

I m = 1k Pmi = N=kI T = m � 1 = (N � k)=kI Rao's resultsCity N M W k m = N=k T = m � 1Tehran 105 65 40 21 5.000 4Isphahan 77 45 32 11 7.000 6Tokyo 124 90 34 50 2.480 1.480Delhi 158 92 66 29 5.448 4.448Calcutta 726 414 312 104 6.980 5.980Waltair 211 123 88 39 5.410 4.410Ahmedabad 133 84 49 29 4.580 3.580Bangalore 307 180 127 55 5.582 4.582



I ResultsI m = 1k Pmi = N=k

I T = m � 1 = (N � k)=kI Rao's resultsCity N M W k m = N=k T = m � 1Tehran 105 65 40 21 5.000 4Isphahan 77 45 32 11 7.000 6Tokyo 124 90 34 50 2.480 1.480Delhi 158 92 66 29 5.448 4.448Calcutta 726 414 312 104 6.980 5.980Waltair 211 123 88 39 5.410 4.410Ahmedabad 133 84 49 29 4.580 3.580Bangalore 307 180 127 55 5.582 4.582



I ResultsI m = 1k Pmi = N=kI T = m � 1 = (N � k)=k

I Rao's resultsCity N M W k m = N=k T = m � 1Tehran 105 65 40 21 5.000 4Isphahan 77 45 32 11 7.000 6Tokyo 124 90 34 50 2.480 1.480Delhi 158 92 66 29 5.448 4.448Calcutta 726 414 312 104 6.980 5.980Waltair 211 123 88 39 5.410 4.410Ahmedabad 133 84 49 29 4.580 3.580Bangalore 307 180 127 55 5.582 4.582



I ResultsI m = 1k Pmi = N=kI T = m � 1 = (N � k)=kI Rao's resultsCity N M W k m = N=k T = m � 1Tehran 105 65 40 21 5.000 4Isphahan 77 45 32 11 7.000 6Tokyo 124 90 34 50 2.480 1.480Delhi 158 92 66 29 5.448 4.448Calcutta 726 414 312 104 6.980 5.980Waltair 211 123 88 39 5.410 4.410Ahmedabad 133 84 49 29 4.580 3.580Bangalore 307 180 127 55 5.582 4.582Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleFisher's example

I R. A., Fisher (1934). The e�ect of methods of ascertainmentupon the estimation of frequencies. Annals Eugenics 6, 13-25.

I Purpose: to study the proportion p of albino children fromnon-albino parents (which can have albino children).I From Medel's laws, p should be 1=4I We do not know if two non-albino parents can have albinochildren!I So Fisher only consider families with albino children.I He only consider families with 5 children, obtaining thefollowing data:



I R. A., Fisher (1934). The e�ect of methods of ascertainmentupon the estimation of frequencies. Annals Eugenics 6, 13-25.I Purpose: to study the proportion p of albino children fromnon-albino parents (which can have albino children).

I From Medel's laws, p should be 1=4I We do not know if two non-albino parents can have albinochildren!I So Fisher only consider families with albino children.I He only consider families with 5 children, obtaining thefollowing data:



I R. A., Fisher (1934). The e�ect of methods of ascertainmentupon the estimation of frequencies. Annals Eugenics 6, 13-25.I Purpose: to study the proportion p of albino children fromnon-albino parents (which can have albino children).I From Medel's laws, p should be 1=4

I We do not know if two non-albino parents can have albinochildren!I So Fisher only consider families with albino children.I He only consider families with 5 children, obtaining thefollowing data:



I R. A., Fisher (1934). The e�ect of methods of ascertainmentupon the estimation of frequencies. Annals Eugenics 6, 13-25.I Purpose: to study the proportion p of albino children fromnon-albino parents (which can have albino children).I From Medel's laws, p should be 1=4I We do not know if two non-albino parents can have albinochildren!

I So Fisher only consider families with albino children.I He only consider families with 5 children, obtaining thefollowing data:



I R. A., Fisher (1934). The e�ect of methods of ascertainmentupon the estimation of frequencies. Annals Eugenics 6, 13-25.I Purpose: to study the proportion p of albino children fromnon-albino parents (which can have albino children).I From Medel's laws, p should be 1=4I We do not know if two non-albino parents can have albinochildren!I So Fisher only consider families with albino children.

I He only consider families with 5 children, obtaining thefollowing data:



I R. A., Fisher (1934). The e�ect of methods of ascertainmentupon the estimation of frequencies. Annals Eugenics 6, 13-25.I Purpose: to study the proportion p of albino children fromnon-albino parents (which can have albino children).I From Medel's laws, p should be 1=4I We do not know if two non-albino parents can have albinochildren!I So Fisher only consider families with albino children.I He only consider families with 5 children, obtaining thefollowing data:Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleFisher data

Number of albino children in the familyN 1 2 3 4 5 Total1 140 80 35 4 0 2592 - 52 12 7 1 723 - - 7 0 0 74 - - - 2 0 25 - - - - 0 0Total 140 132 54 13 1 340I N=Number of albino children in the sample.

I N�otice that we have 340 families sampled from 432 di�erentalbino children.


Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleFisher data

Number of albino children in the familyN 1 2 3 4 5 Total1 140 80 35 4 0 2592 - 52 12 7 1 723 - - 7 0 0 74 - - - 2 0 25 - - - - 0 0Total 140 132 54 13 1 340I N=Number of albino children in the sample.I N�otice that we have 340 families sampled from 432 di�erentalbino children.


Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleSolution 1

I What to do with these data?

I If X1; :::;Xn is a sampe of size n = 340 from aBinomialB(k = 5; p = 1=4),I p can be estimated asbp1 =

Pni=1 Xi5n = 140 + 2 � 132 + :::5 � 340 = 6231700 = 0:3665I with variance

�2(bp1) = p(1� p)5n = 0:25 � 0:751700 = 0:0001:I This gives 2�(bp1) ' 0:021 and we reject p = 0:25.



I What to do with these data?I If X1; :::;Xn is a sampe of size n = 340 from aBinomialB(k = 5; p = 1=4),

I p can be estimated asbp1 =





I What to do with these data?I If X1; :::;Xn is a sampe of size n = 340 from aBinomialB(k = 5; p = 1=4),I p can be estimated asbp1 =

Pni=1 Xi5n = 140 + 2 � 132 + :::5 � 340 = 6231700 = 0:3665

I with variance�2(bp1) = p(1� p)5n = 0:25 � 0:751700 = 0:0001:

I This gives 2�(bp1) ' 0:021 and we reject p = 0:25.





�2(bp1) = p(1� p)5n = 0:25 � 0:751700 = 0:0001:

I This gives 2�(bp1) ' 0:021 and we reject p = 0:25.


Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleSolution 1bis

I If we use the families several times then

I the sample size is n = 432 and p is estimated asbp1 =

Pni=1 Xi5n = 140 + 2 � 184 + :::5 � 432 = 0:399I This also leads to reject p = 0:25.



I If we use the families several times thenI the sample size is n = 432 and p is estimated asbp1 =

Pni=1 Xi5n = 140 + 2 � 184 + :::5 � 432 = 0:399

I This also leads to reject p = 0:25.



I If we use the families several times thenI the sample size is n = 432 and p is estimated asbp1 =

Pni=1 Xi5n = 140 + 2 � 184 + :::5 � 432 = 0:399I This also leads to reject p = 0:25.



I The families with 0 albino children cannot appear in thesample.

I Thus, we might think in a censored sample with w(x) = 1 forx 6= 0 and w(0) = 0:I Then p�(x) = p(x)=(1� q5); where p(x) � BinomialB(5; 1=4)I Then the MLE satis�esp1� q5 = Pni=1 Xi5n ;

I which gives bp2 = 0:3085 (bp2 = 0:35 with the repeatedfamilies).I In both cases we reject p = 1=4:



I The families with 0 albino children cannot appear in thesample.I Thus, we might think in a censored sample with w(x) = 1 forx 6= 0 and w(0) = 0:

I Then p�(x) = p(x)=(1� q5); where p(x) � BinomialB(5; 1=4)I Then the MLE satis�esp1� q5 = Pni=1 Xi5n ;




I The families with 0 albino children cannot appear in thesample.I Thus, we might think in a censored sample with w(x) = 1 forx 6= 0 and w(0) = 0:I Then p�(x) = p(x)=(1� q5); where p(x) � BinomialB(5; 1=4)

I Then the MLE satis�esp1� q5 = Pni=1 Xi5n ;




I The families with 0 albino children cannot appear in thesample.I Thus, we might think in a censored sample with w(x) = 1 forx 6= 0 and w(0) = 0:I Then p�(x) = p(x)=(1� q5); where p(x) � BinomialB(5; 1=4)I Then the MLE satis�esp1� q5 = Pni=1 Xi5n ;





I which gives bp2 = 0:3085 (bp2 = 0:35 with the repeatedfamilies).

I In both cases we reject p = 1=4:




I which gives bp2 = 0:3085 (bp2 = 0:35 with the repeatedfamilies).I In both cases we reject p = 1=4:Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix De�nitionRao's exampleFisher's exampleSolution 3 (the correct one)

I Note that the sampling probability of a family with x albinochildren is proportional to x .

I Then Xi � length biased Binomial.I That is, Xi � 1 �Binomial B(4; 1=4)I Then, using the repeated families p is estimated asbp3 =

Pni=1(Xi � 1)4n = 1 � 184 + 2 � 80 + :::4 � 432 = 0:2488I The variance satis�es 2�(bp3) ' 0:0208, which is consistentwith p = 1=4.I Notice that if we do not use the repeated families the p isunderestimated as

bp4 =Pni=1(Xi � 1)4n = 1 � 132 + 2 � 54 + :::4 � 340 = 0:2080



I Note that the sampling probability of a family with x albinochildren is proportional to x .I Then Xi � length biased Binomial.

I That is, Xi � 1 �Binomial B(4; 1=4)I Then, using the repeated families p is estimated asbp3 =


bp4 =Pni=1(Xi � 1)4n = 1 � 132 + 2 � 54 + :::4 � 340 = 0:2080



I Note that the sampling probability of a family with x albinochildren is proportional to x .I Then Xi � length biased Binomial.I That is, Xi � 1 �Binomial B(4; 1=4)

I Then, using the repeated families p is estimated asbp3 =


bp4 =Pni=1(Xi � 1)4n = 1 � 132 + 2 � 54 + :::4 � 340 = 0:2080



I Note that the sampling probability of a family with x albinochildren is proportional to x .I Then Xi � length biased Binomial.I That is, Xi � 1 �Binomial B(4; 1=4)I Then, using the repeated families p is estimated asbp3 =

Pni=1(Xi � 1)4n = 1 � 184 + 2 � 80 + :::4 � 432 = 0:2488

I The variance satis�es 2�(bp3) ' 0:0208, which is consistentwith p = 1=4.I Notice that if we do not use the repeated families the p isunderestimated asbp4 =

Pni=1(Xi � 1)4n = 1 � 132 + 2 � 54 + :::4 � 340 = 0:2080




Pni=1(Xi � 1)4n = 1 � 184 + 2 � 80 + :::4 � 432 = 0:2488I The variance satis�es 2�(bp3) ' 0:0208, which is consistentwith p = 1=4.

I Notice that if we do not use the repeated families the p isunderestimated asbp4 =

Pni=1(Xi � 1)4n = 1 � 132 + 2 � 54 + :::4 � 340 = 0:2080





bp4 =Pni=1(Xi � 1)4n = 1 � 132 + 2 � 54 + :::4 � 340 = 0:2080


Biased samplesRenewal processesHow to detect biased samples?Appendix Waiting time paradoxEquilibrium distributionWaiting time paradox

Figure: If a passenger arrives at a bus-stop at some random point andthe interval time between the buses is 20 min, what is the mean waitingtime until the next bus?Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)


I R.C. Gupta 1979. Waiting time paradox and size biasedsampling. Communications in Statistics, Theory and MethodsA8 (6), 601-607.

I Let us assume that the buses pass every 20min. and that wedo not know the time table:j � � ��j � ��j � ��j � ��j � ��j � ��j

I Then the waiting time T should be Uniform (0; 20)I Then the expected waiting time should beE (T ) = 20=2 = 10min :I We know that this is not true!



I R.C. Gupta 1979. Waiting time paradox and size biasedsampling. Communications in Statistics, Theory and MethodsA8 (6), 601-607.I Let us assume that the buses pass every 20min. and that wedo not know the time table:j � � ��j � ��j � ��j � ��j � ��j � ��j

I Then the waiting time T should be Uniform (0; 20)I Then the expected waiting time should beE (T ) = 20=2 = 10min :I We know that this is not true!




I Then the waiting time T should be Uniform (0; 20)

I Then the expected waiting time should beE (T ) = 20=2 = 10min :I We know that this is not true!




I Then the waiting time T should be Uniform (0; 20)I Then the expected waiting time should beE (T ) = 20=2 = 10min :

I We know that this is not true!




I Then the waiting time T should be Uniform (0; 20)I Then the expected waiting time should beE (T ) = 20=2 = 10min :I We know that this is not true!Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)


I The real times are:j �� j �� j�� j��j ��j�� j

I Then the time between buses is a random variable X .I Let us assume that � = E (X ) = 20min :I Then T �Uniforme (0;X )I The waiting time should be T = UX ; where U �Uniform(0; 1).I Then the expected waiting time should beE (T ) = E (UX ) = E (X )=2 = 10minI We know that this is not true!




I Then the time between buses is a random variable X .

I Let us assume that � = E (X ) = 20min :I Then T �Uniforme (0;X )I The waiting time should be T = UX ; where U �Uniform(0; 1).I Then the expected waiting time should beE (T ) = E (UX ) = E (X )=2 = 10minI We know that this is not true!




I Then the time between buses is a random variable X .I Let us assume that � = E (X ) = 20min :

I Then T �Uniforme (0;X )I The waiting time should be T = UX ; where U �Uniform(0; 1).I Then the expected waiting time should beE (T ) = E (UX ) = E (X )=2 = 10minI We know that this is not true!




I Then the time between buses is a random variable X .I Let us assume that � = E (X ) = 20min :I Then T �Uniforme (0;X )

I The waiting time should be T = UX ; where U �Uniform(0; 1).I Then the expected waiting time should beE (T ) = E (UX ) = E (X )=2 = 10minI We know that this is not true!




I Then the time between buses is a random variable X .I Let us assume that � = E (X ) = 20min :I Then T �Uniforme (0;X )I The waiting time should be T = UX ; where U �Uniform(0; 1).

I Then the expected waiting time should beE (T ) = E (UX ) = E (X )=2 = 10minI We know that this is not true!




I Then the time between buses is a random variable X .I Let us assume that � = E (X ) = 20min :I Then T �Uniforme (0;X )I The waiting time should be T = UX ; where U �Uniform(0; 1).I Then the expected waiting time should beE (T ) = E (UX ) = E (X )=2 = 10min

I We know that this is not true!




I Then the time between buses is a random variable X .I Let us assume that � = E (X ) = 20min :I Then T �Uniforme (0;X )I The waiting time should be T = UX ; where U �Uniform(0; 1).I Then the expected waiting time should beE (T ) = E (UX ) = E (X )=2 = 10minI We know that this is not true!Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)



I If X1;X2; :::;Xn are the times between buses, the theprobability of have a time Xi is proportional to Xi .I Then T �Uniform (0;X �); where X � is the length biased r.v.E (X �) = Z 1

0 xf �(x)dx = Z 10 x xf (x)� dx = E (X 2)E (X ) = �+ �2�

I That is E (T ) = E (X �=2) = 10 + �2=(20) > 10I We only have E (T ) = 10 if �2 = 0!I It is very important the regularity!




I If X1;X2; :::;Xn are the times between buses, the theprobability of have a time Xi is proportional to Xi .

I Then T �Uniform (0;X �); where X � is the length biased r.v.E (X �) = Z 1








I That is E (T ) = E (X �=2) = 10 + �2=(20) > 10

I We only have E (T ) = 10 if �2 = 0!I It is very important the regularity!






I That is E (T ) = E (X �=2) = 10 + �2=(20) > 10I We only have E (T ) = 10 if �2 = 0!

I It is very important the regularity!






I That is E (T ) = E (X �=2) = 10 + �2=(20) > 10I We only have E (T ) = 10 if �2 = 0!I It is very important the regularity!Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix Waiting time paradoxEquilibrium distributionExponential case

I In particular, if X � Exp(� = 20min)

I E (X �) = �+ �2=� = �+ �2=� = 2�I E (T ) = E (X �=2) = E (X )!!I Paradox: If the expected time between buses is 20 min., wehave to wait 20 min.!I Similar results are obtained in renewal processes (with randominspections).



I In particular, if X � Exp(� = 20min)I E (X �) = �+ �2=� = �+ �2=� = 2�

I E (T ) = E (X �=2) = E (X )!!I Paradox: If the expected time between buses is 20 min., wehave to wait 20 min.!I Similar results are obtained in renewal processes (with randominspections).



I In particular, if X � Exp(� = 20min)I E (X �) = �+ �2=� = �+ �2=� = 2�I E (T ) = E (X �=2) = E (X )!!

I Paradox: If the expected time between buses is 20 min., wehave to wait 20 min.!I Similar results are obtained in renewal processes (with randominspections).



I In particular, if X � Exp(� = 20min)I E (X �) = �+ �2=� = �+ �2=� = 2�I E (T ) = E (X �=2) = E (X )!!I Paradox: If the expected time between buses is 20 min., wehave to wait 20 min.!

I Similar results are obtained in renewal processes (with randominspections).



I In particular, if X � Exp(� = 20min)I E (X �) = �+ �2=� = �+ �2=� = 2�I E (T ) = E (X �=2) = E (X )!!I Paradox: If the expected time between buses is 20 min., wehave to wait 20 min.!I Similar results are obtained in renewal processes (with randominspections).


Biased samplesRenewal processesHow to detect biased samples?Appendix Waiting time paradoxEquilibrium distributionGeneral solution

I When a unit fails, it is replaced by a similar onej �� j �� j�� j��j ��j�� j

I The unit lifetimes X1;X2; ::: are i.i.d. from XI We do random inspections.I The forward (or backward) time from a sample point isT = UX ; where U �Uniform (0; 1) (X and U areindependent).I This is not true!




I The unit lifetimes X1;X2; ::: are i.i.d. from X

I We do random inspections.I The forward (or backward) time from a sample point isT = UX ; where U �Uniform (0; 1) (X and U areindependent).I This is not true!




I The unit lifetimes X1;X2; ::: are i.i.d. from XI We do random inspections.

I The forward (or backward) time from a sample point isT = UX ; where U �Uniform (0; 1) (X and U areindependent).I This is not true!




I The unit lifetimes X1;X2; ::: are i.i.d. from XI We do random inspections.I The forward (or backward) time from a sample point isT = UX ; where U �Uniform (0; 1) (X and U areindependent).

I This is not true!




I The unit lifetimes X1;X2; ::: are i.i.d. from XI We do random inspections.I The forward (or backward) time from a sample point isT = UX ; where U �Uniform (0; 1) (X and U areindependent).I This is not true!



I The correct solution is T = UX �; and hencef (x ; u) = f �(x) = xf (x)� ; 0 < u < 1; x > 0

I If FT (t) = Pr(T > t) = Pr(UX � > t), t > 0,FT (t) = Z 1

tZ 1t=x

xf (x)� dudx = Z 1t

(x � t)f (x)� dx = Z 1t

F (x)� dxI Thus,

fT (t) = F 0T (t) = F (t)� = 1� F (t)f (t) f (t)� = w(t) f (t)� ; t > 0w(t) = 1� F (t)f (t) = 1h(t) ; where h(t) = f (t)1� F (t) is the hazard rate





tZ 1t=x


(x � t)f (x)� dx = Z 1t

F (x)� dx

I Thus,fT (t) = F 0T (t) = F (t)� = 1� F (t)f (t) f (t)� = w(t) f (t)� ; t > 0w(t) = 1� F (t)f (t) = 1h(t) ; where h(t) = f (t)1� F (t) is the hazard rate





tZ 1t=x


(x � t)f (x)� dx = Z 1t

F (x)� dxI Thus,

fT (t) = F 0T (t) = F (t)� = 1� F (t)f (t) f (t)� = w(t) f (t)� ; t > 0w(t) = 1� F (t)f (t) = 1h(t) ; where h(t) = f (t)1� F (t) is the hazard rate


Biased samplesRenewal processesHow to detect biased samples?Appendix Waiting time paradoxEquilibrium distributionWaiting time solution

I If X �, T = UX � , t > 0,FT (t) = Z 1

tF (x)� dx = Z 1

texp(�x=�)� dx = exp(�t=�)

I That is, T �Exponential: X =d UX �.I Actually, the exponential model is the unique model such thatX =d UX � (or X =d T ).I The distribution of T is called the equilibrium distribution andvery interesting properties.I For example,hT (t) = fT (t)FT (t) = FT (t)R1t FT (x)dx = 1m(t)

where m(t) = E (X � tjX > t) in the mean residual life.



I If X �, T = UX � , t > 0,FT (t) = Z 1

tF (x)� dx = Z 1


I That is, T �Exponential: X =d UX �.

I Actually, the exponential model is the unique model such thatX =d UX � (or X =d T ).I The distribution of T is called the equilibrium distribution andvery interesting properties.I For example,hT (t) = fT (t)FT (t) = FT (t)R1t FT (x)dx = 1m(t)




I If X �, T = UX � , t > 0,FT (t) = Z 1

tF (x)� dx = Z 1


I That is, T �Exponential: X =d UX �.I Actually, the exponential model is the unique model such thatX =d UX � (or X =d T ).

I The distribution of T is called the equilibrium distribution andvery interesting properties.I For example,hT (t) = fT (t)FT (t) = FT (t)R1t FT (x)dx = 1m(t)




I If X �, T = UX � , t > 0,FT (t) = Z 1

tF (x)� dx = Z 1


I That is, T �Exponential: X =d UX �.I Actually, the exponential model is the unique model such thatX =d UX � (or X =d T ).I The distribution of T is called the equilibrium distribution andvery interesting properties.

I For example,hT (t) = fT (t)FT (t) = FT (t)R1t FT (x)dx = 1m(t)




I If X �, T = UX � , t > 0,FT (t) = Z 1

tF (x)� dx = Z 1


I That is, T �Exponential: X =d UX �.I Actually, the exponential model is the unique model such thatX =d UX � (or X =d T ).I The distribution of T is called the equilibrium distribution andvery interesting properties.I For example,hT (t) = fT (t)FT (t) = FT (t)R1t FT (x)dx = 1m(t)

where m(t) = E (X � tjX > t) in the mean residual life.Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix Mean sojourn time per touristHow to be a rich man?How to detect biased samples?

I In Fisher and Rao examples the results do not �t to theexpected values.

I In the waiting time paradox the results do not coindice withour experience.I This example is based on two surveys to study the meansojourn time per tourist in Morocco (INSEA, 1966).I G.P. Patil 1981. Proceedings of the Indian Statistical InstituteJubilee International Conference on Statistics: Applicationsand New Directions, 478-503)I A sample at the border stations: n=3000, mean=9.0 daysI A sample at the hotels: n=12321, mean=17.8 daysI The results are very di�erent!I The second sample was discarded.



I In Fisher and Rao examples the results do not �t to theexpected values.I In the waiting time paradox the results do not coindice withour experience.

I This example is based on two surveys to study the meansojourn time per tourist in Morocco (INSEA, 1966).I G.P. Patil 1981. Proceedings of the Indian Statistical InstituteJubilee International Conference on Statistics: Applicationsand New Directions, 478-503)I A sample at the border stations: n=3000, mean=9.0 daysI A sample at the hotels: n=12321, mean=17.8 daysI The results are very di�erent!I The second sample was discarded.



I In Fisher and Rao examples the results do not �t to theexpected values.I In the waiting time paradox the results do not coindice withour experience.I This example is based on two surveys to study the meansojourn time per tourist in Morocco (INSEA, 1966).I G.P. Patil 1981. Proceedings of the Indian Statistical InstituteJubilee International Conference on Statistics: Applicationsand New Directions, 478-503)

I A sample at the border stations: n=3000, mean=9.0 daysI A sample at the hotels: n=12321, mean=17.8 daysI The results are very di�erent!I The second sample was discarded.



I In Fisher and Rao examples the results do not �t to theexpected values.I In the waiting time paradox the results do not coindice withour experience.I This example is based on two surveys to study the meansojourn time per tourist in Morocco (INSEA, 1966).I G.P. Patil 1981. Proceedings of the Indian Statistical InstituteJubilee International Conference on Statistics: Applicationsand New Directions, 478-503)I A sample at the border stations: n=3000, mean=9.0 days

I A sample at the hotels: n=12321, mean=17.8 daysI The results are very di�erent!I The second sample was discarded.



I In Fisher and Rao examples the results do not �t to theexpected values.I In the waiting time paradox the results do not coindice withour experience.I This example is based on two surveys to study the meansojourn time per tourist in Morocco (INSEA, 1966).I G.P. Patil 1981. Proceedings of the Indian Statistical InstituteJubilee International Conference on Statistics: Applicationsand New Directions, 478-503)I A sample at the border stations: n=3000, mean=9.0 daysI A sample at the hotels: n=12321, mean=17.8 days

I The results are very di�erent!I The second sample was discarded.



I In Fisher and Rao examples the results do not �t to theexpected values.I In the waiting time paradox the results do not coindice withour experience.I This example is based on two surveys to study the meansojourn time per tourist in Morocco (INSEA, 1966).I G.P. Patil 1981. Proceedings of the Indian Statistical InstituteJubilee International Conference on Statistics: Applicationsand New Directions, 478-503)I A sample at the border stations: n=3000, mean=9.0 daysI A sample at the hotels: n=12321, mean=17.8 daysI The results are very di�erent!

I The second sample was discarded.



I In Fisher and Rao examples the results do not �t to theexpected values.I In the waiting time paradox the results do not coindice withour experience.I This example is based on two surveys to study the meansojourn time per tourist in Morocco (INSEA, 1966).I G.P. Patil 1981. Proceedings of the Indian Statistical InstituteJubilee International Conference on Statistics: Applicationsand New Directions, 478-503)I A sample at the border stations: n=3000, mean=9.0 daysI A sample at the hotels: n=12321, mean=17.8 daysI The results are very di�erent!I The second sample was discarded.Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix Mean sojourn time per touristHow to be a rich man?Patil's Solution

I The sample at the hotels is length biased !

I A tourist staying 6 days has double sampling probability thana tourist staying 3 days.I Hence E (X ) ' 9:0 < E (X �) ' 17:8I Actually E (X �) ' 2E (X ) might indicate that X isExponential.I Correct estimation E (X ) ' 17:8=2 = 8: 9I Similar examples in other �elds.



I The sample at the hotels is length biased !I A tourist staying 6 days has double sampling probability thana tourist staying 3 days.

I Hence E (X ) ' 9:0 < E (X �) ' 17:8I Actually E (X �) ' 2E (X ) might indicate that X isExponential.I Correct estimation E (X ) ' 17:8=2 = 8: 9I Similar examples in other �elds.



I The sample at the hotels is length biased !I A tourist staying 6 days has double sampling probability thana tourist staying 3 days.I Hence E (X ) ' 9:0 < E (X �) ' 17:8

I Actually E (X �) ' 2E (X ) might indicate that X isExponential.I Correct estimation E (X ) ' 17:8=2 = 8: 9I Similar examples in other �elds.



I The sample at the hotels is length biased !I A tourist staying 6 days has double sampling probability thana tourist staying 3 days.I Hence E (X ) ' 9:0 < E (X �) ' 17:8I Actually E (X �) ' 2E (X ) might indicate that X isExponential.

I Correct estimation E (X ) ' 17:8=2 = 8: 9I Similar examples in other �elds.



I The sample at the hotels is length biased !I A tourist staying 6 days has double sampling probability thana tourist staying 3 days.I Hence E (X ) ' 9:0 < E (X �) ' 17:8I Actually E (X �) ' 2E (X ) might indicate that X isExponential.I Correct estimation E (X ) ' 17:8=2 = 8: 9

I Similar examples in other �elds.



I The sample at the hotels is length biased !I A tourist staying 6 days has double sampling probability thana tourist staying 3 days.I Hence E (X ) ' 9:0 < E (X �) ' 17:8I Actually E (X �) ' 2E (X ) might indicate that X isExponential.I Correct estimation E (X ) ' 17:8=2 = 8: 9I Similar examples in other �elds.


Biased samplesRenewal processesHow to detect biased samples?Appendix Mean sojourn time per touristHow to be a rich man?What to do?

I The best solution is to use all the information available!

I X1; ::::;Xn is a sample from X � Exp(�).I Y1; ::::;Ym is a sample from X � � Exp�(�).I The MLE (exponential) is:b� = 1n + 2m

0@ nX

i=1 Xi +mXj=1 Yj

1A

I It is unbiased since E (Xi ) = � y E (Yj) = 2�I With varianceVar(b�) = �2n + 2mI It is the UMVUE.



I The best solution is to use all the information available!I X1; ::::;Xn is a sample from X � Exp(�).

I Y1; ::::;Ym is a sample from X � � Exp�(�).I The MLE (exponential) is:b� = 1n + 2m

0@ nX

i=1 Xi +mXj=1 Yj

1A




I The best solution is to use all the information available!I X1; ::::;Xn is a sample from X � Exp(�).I Y1; ::::;Ym is a sample from X � � Exp�(�).

I The MLE (exponential) is:b� = 1n + 2m

0@ nX

i=1 Xi +mXj=1 Yj

1A




I The best solution is to use all the information available!I X1; ::::;Xn is a sample from X � Exp(�).I Y1; ::::;Ym is a sample from X � � Exp�(�).I The MLE (exponential) is:b� = 1n + 2m

0@ nX

i=1 Xi +mXj=1 Yj

1A





0@ nX

i=1 Xi +mXj=1 Yj

1A

I It is unbiased since E (Xi ) = � y E (Yj) = 2�

I With varianceVar(b�) = �2n + 2mI It is the UMVUE.




0@ nX

i=1 Xi +mXj=1 Yj

1A

I It is unbiased since E (Xi ) = � y E (Yj) = 2�I With varianceVar(b�) = �2n + 2m

I It is the UMVUE.




0@ nX

i=1 Xi +mXj=1 Yj

1A

I It is unbiased since E (Xi ) = � y E (Yj) = 2�I With varianceVar(b�) = �2n + 2mI It is the UMVUE. Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix Mean sojourn time per touristHow to be a rich man?The best estimation

I The best estimation isb� = 13000 + 2 � 12321(3000 � 9 + 12321 � 17:8) = 8: 91;

I with variance 2�(b�) ' 0:11I First sample b� = 9:0I Second sample b� = 8:9




I with variance 2�(b�) ' 0:11

I First sample b� = 9:0I Second sample b� = 8:9




I with variance 2�(b�) ' 0:11I First sample b� = 9:0

I Second sample b� = 8:9




I with variance 2�(b�) ' 0:11I First sample b� = 9:0I Second sample b� = 8:9


Biased samplesRenewal processesHow to detect biased samples?Appendix Mean sojourn time per touristHow to be a rich man?General solution in the exponencial case

I What to do the next time ?

I Which sample is the best one?I Notice that the estimator from the second sample has lessvariance.I The Fisher information for n = m = 1 areI �(�) = 2=�2I (�) = 1=�2

I Each data in the second sample has double information!I If the bias is w(x) = xk ; the Fisher information is increasingin k .I Other models, see Navarro et al. (2001, Biom. J. 43).



I What to do the next time ?I Which sample is the best one?

I Notice that the estimator from the second sample has lessvariance.I The Fisher information for n = m = 1 areI �(�) = 2=�2I (�) = 1=�2




I What to do the next time ?I Which sample is the best one?I Notice that the estimator from the second sample has lessvariance.

I The Fisher information for n = m = 1 areI �(�) = 2=�2I (�) = 1=�2




I What to do the next time ?I Which sample is the best one?I Notice that the estimator from the second sample has lessvariance.I The Fisher information for n = m = 1 areI �(�) = 2=�2I (�) = 1=�2





I Each data in the second sample has double information!

I If the bias is w(x) = xk ; the Fisher information is increasingin k .I Other models, see Navarro et al. (2001, Biom. J. 43).




I Each data in the second sample has double information!I If the bias is w(x) = xk ; the Fisher information is increasingin k .

I Other models, see Navarro et al. (2001, Biom. J. 43).




I Each data in the second sample has double information!I If the bias is w(x) = xk ; the Fisher information is increasingin k .I Other models, see Navarro et al. (2001, Biom. J. 43).Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix Mean sojourn time per touristHow to be a rich man?DeGroot's example

I G.P. Patil (1991). Encountered data, Statistical EcologyEnvironmental Statistics and weighted distribution methods.Environmetrics 2(4), 377-423.

I Discussion by M.H. DeGroot (Profesor Carnegie MellonUniversity, Pittsburgh).I How to predict the stock market behavior?I We send 128 letters, 64 saying:I \I'm an expert analyst and I have a model to predict the thestock market behavior. To show that I inform you fro FREEthat the stocks of the company SOME are going to go UPthis week.".I The other 64 letters say: \... to go DOWN this week".



I G.P. Patil (1991). Encountered data, Statistical EcologyEnvironmental Statistics and weighted distribution methods.Environmetrics 2(4), 377-423.I Discussion by M.H. DeGroot (Profesor Carnegie MellonUniversity, Pittsburgh).

I How to predict the stock market behavior?I We send 128 letters, 64 saying:I \I'm an expert analyst and I have a model to predict the thestock market behavior. To show that I inform you fro FREEthat the stocks of the company SOME are going to go UPthis week.".I The other 64 letters say: \... to go DOWN this week".



I G.P. Patil (1991). Encountered data, Statistical EcologyEnvironmental Statistics and weighted distribution methods.Environmetrics 2(4), 377-423.I Discussion by M.H. DeGroot (Profesor Carnegie MellonUniversity, Pittsburgh).I How to predict the stock market behavior?

I We send 128 letters, 64 saying:I \I'm an expert analyst and I have a model to predict the thestock market behavior. To show that I inform you fro FREEthat the stocks of the company SOME are going to go UPthis week.".I The other 64 letters say: \... to go DOWN this week".



I G.P. Patil (1991). Encountered data, Statistical EcologyEnvironmental Statistics and weighted distribution methods.Environmetrics 2(4), 377-423.I Discussion by M.H. DeGroot (Profesor Carnegie MellonUniversity, Pittsburgh).I How to predict the stock market behavior?I We send 128 letters, 64 saying:

I \I'm an expert analyst and I have a model to predict the thestock market behavior. To show that I inform you fro FREEthat the stocks of the company SOME are going to go UPthis week.".I The other 64 letters say: \... to go DOWN this week".



I G.P. Patil (1991). Encountered data, Statistical EcologyEnvironmental Statistics and weighted distribution methods.Environmetrics 2(4), 377-423.I Discussion by M.H. DeGroot (Profesor Carnegie MellonUniversity, Pittsburgh).I How to predict the stock market behavior?I We send 128 letters, 64 saying:I \I'm an expert analyst and I have a model to predict the thestock market behavior. To show that I inform you fro FREEthat the stocks of the company SOME are going to go UPthis week.".

I The other 64 letters say: \... to go DOWN this week".



I G.P. Patil (1991). Encountered data, Statistical EcologyEnvironmental Statistics and weighted distribution methods.Environmetrics 2(4), 377-423.I Discussion by M.H. DeGroot (Profesor Carnegie MellonUniversity, Pittsburgh).I How to predict the stock market behavior?I We send 128 letters, 64 saying:I \I'm an expert analyst and I have a model to predict the thestock market behavior. To show that I inform you fro FREEthat the stocks of the company SOME are going to go UPthis week.".I The other 64 letters say: \... to go DOWN this week".Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)


I The next week we will send similar letter but only to thepeople (64) with the correct predictions saying:

I \Last week I sent you a correct predictions. To show you thatmy model does not fail I send you another correct predictionfor FREE this week: the stocks of the company SOME aregoing to go UP (DOWN)".I We repeat this process 7 weeks.I Finally we sent the following letter:I \Well I think that I have show you that my model does notfail. Now if you want to know the next prediction you have topay 10.000$".



I The next week we will send similar letter but only to thepeople (64) with the correct predictions saying:I \Last week I sent you a correct predictions. To show you thatmy model does not fail I send you another correct predictionfor FREE this week: the stocks of the company SOME aregoing to go UP (DOWN)".

I We repeat this process 7 weeks.I Finally we sent the following letter:I \Well I think that I have show you that my model does notfail. Now if you want to know the next prediction you have topay 10.000$".



I The next week we will send similar letter but only to thepeople (64) with the correct predictions saying:I \Last week I sent you a correct predictions. To show you thatmy model does not fail I send you another correct predictionfor FREE this week: the stocks of the company SOME aregoing to go UP (DOWN)".I We repeat this process 7 weeks.

I Finally we sent the following letter:I \Well I think that I have show you that my model does notfail. Now if you want to know the next prediction you have topay 10.000$".



I The next week we will send similar letter but only to thepeople (64) with the correct predictions saying:I \Last week I sent you a correct predictions. To show you thatmy model does not fail I send you another correct predictionfor FREE this week: the stocks of the company SOME aregoing to go UP (DOWN)".I We repeat this process 7 weeks.I Finally we sent the following letter:

I \Well I think that I have show you that my model does notfail. Now if you want to know the next prediction you have topay 10.000$".



I The next week we will send similar letter but only to thepeople (64) with the correct predictions saying:I \Last week I sent you a correct predictions. To show you thatmy model does not fail I send you another correct predictionfor FREE this week: the stocks of the company SOME aregoing to go UP (DOWN)".I We repeat this process 7 weeks.I Finally we sent the following letter:I \Well I think that I have show you that my model does notfail. Now if you want to know the next prediction you have topay 10.000$".Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix Mean sojourn time per touristHow to be a rich man?Solution of DeGroot's example

I We have a sample X1; :::;X7 from a Bernoulli B(p) with aprobability p of a correct prediction Xi = 1 and a estimationbp = 7=7 = 1:

I But, what is the probability of a value Xi appear in thesample?I Clearly, it is proportional to Xi !I That is we have a sample from the length biased r.v. X � withp�(x) = xp(x)=�, x = 0; 1, that is, X � = 1.



I We have a sample X1; :::;X7 from a Bernoulli B(p) with aprobability p of a correct prediction Xi = 1 and a estimationbp = 7=7 = 1:I But, what is the probability of a value Xi appear in thesample?

I Clearly, it is proportional to Xi !I That is we have a sample from the length biased r.v. X � withp�(x) = xp(x)=�, x = 0; 1, that is, X � = 1.



I We have a sample X1; :::;X7 from a Bernoulli B(p) with aprobability p of a correct prediction Xi = 1 and a estimationbp = 7=7 = 1:I But, what is the probability of a value Xi appear in thesample?I Clearly, it is proportional to Xi !

I That is we have a sample from the length biased r.v. X � withp�(x) = xp(x)=�, x = 0; 1, that is, X � = 1.



I We have a sample X1; :::;X7 from a Bernoulli B(p) with aprobability p of a correct prediction Xi = 1 and a estimationbp = 7=7 = 1:I But, what is the probability of a value Xi appear in thesample?I Clearly, it is proportional to Xi !I That is we have a sample from the length biased r.v. X � withp�(x) = xp(x)=�, x = 0; 1, that is, X � = 1.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesConclusions

I We have to think about the selection methods in a sample!

I They can be biased but with a known bias.I If we have two di�erent samples (with a known bias), the bestsolution is always to use both together. We need to changethe classical estimators.I With a biased sample, we can obtain results as good as (oreven better) that an unbiased sample. We need to change theclassical estimators.I If we have to choose, we should use the sample (biased ornot) with the highest information about the parameter.



I We have to think about the selection methods in a sample!I They can be biased but with a known bias.

I If we have two di�erent samples (with a known bias), the bestsolution is always to use both together. We need to changethe classical estimators.I With a biased sample, we can obtain results as good as (oreven better) that an unbiased sample. We need to change theclassical estimators.I If we have to choose, we should use the sample (biased ornot) with the highest information about the parameter.



I We have to think about the selection methods in a sample!I They can be biased but with a known bias.I If we have two di�erent samples (with a known bias), the bestsolution is always to use both together. We need to changethe classical estimators.

I With a biased sample, we can obtain results as good as (oreven better) that an unbiased sample. We need to change theclassical estimators.I If we have to choose, we should use the sample (biased ornot) with the highest information about the parameter.



I We have to think about the selection methods in a sample!I They can be biased but with a known bias.I If we have two di�erent samples (with a known bias), the bestsolution is always to use both together. We need to changethe classical estimators.I With a biased sample, we can obtain results as good as (oreven better) that an unbiased sample. We need to change theclassical estimators.

I If we have to choose, we should use the sample (biased ornot) with the highest information about the parameter.



I We have to think about the selection methods in a sample!I They can be biased but with a known bias.I If we have two di�erent samples (with a known bias), the bestsolution is always to use both together. We need to changethe classical estimators.I With a biased sample, we can obtain results as good as (oreven better) that an unbiased sample. We need to change theclassical estimators.I If we have to choose, we should use the sample (biased ornot) with the highest information about the parameter.Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Guillamon, Navarro and Ruiz (1998). Kernel densityestimation using weighted data. Comm. Statist. Theory andMethods 27, 2123-2135.

I Navarro, Ruiz and del Aguila (2001). Parametric estimationfrom weighted samples. Biometrical J. 43, 297-311.I Navarro, Ruiz and del Aguila (2003). How to detect biasedsamples?. Biometrical J. 44, 742-763.I Pakes, Navarro, Ruiz and del Aguila (2003). Characterizationsusing weighted distributions. JSPI 116, 389-420.I Navarro, Ruiz and del Aguila (2006). Multivariate weighteddistributions. Statist. 40 (1), 51-54.I Pakes and Navarro (2007). Distributional characterizationsthrough scaling relations. Australian and New Zealand J.Statist. 49 (2), 115-135.I Navarro and Sarabia (2010). Alternative de�nitions ofbivariate equilibrium distributions. JSPI 140, 2046-2056.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Guillamon, Navarro and Ruiz (1998). Kernel densityestimation using weighted data. Comm. Statist. Theory andMethods 27, 2123-2135.I Navarro, Ruiz and del Aguila (2001). Parametric estimationfrom weighted samples. Biometrical J. 43, 297-311.

I Navarro, Ruiz and del Aguila (2003). How to detect biasedsamples?. Biometrical J. 44, 742-763.I Pakes, Navarro, Ruiz and del Aguila (2003). Characterizationsusing weighted distributions. JSPI 116, 389-420.I Navarro, Ruiz and del Aguila (2006). Multivariate weighteddistributions. Statist. 40 (1), 51-54.I Pakes and Navarro (2007). Distributional characterizationsthrough scaling relations. Australian and New Zealand J.Statist. 49 (2), 115-135.I Navarro and Sarabia (2010). Alternative de�nitions ofbivariate equilibrium distributions. JSPI 140, 2046-2056.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Guillamon, Navarro and Ruiz (1998). Kernel densityestimation using weighted data. Comm. Statist. Theory andMethods 27, 2123-2135.I Navarro, Ruiz and del Aguila (2001). Parametric estimationfrom weighted samples. Biometrical J. 43, 297-311.I Navarro, Ruiz and del Aguila (2003). How to detect biasedsamples?. Biometrical J. 44, 742-763.

I Pakes, Navarro, Ruiz and del Aguila (2003). Characterizationsusing weighted distributions. JSPI 116, 389-420.I Navarro, Ruiz and del Aguila (2006). Multivariate weighteddistributions. Statist. 40 (1), 51-54.I Pakes and Navarro (2007). Distributional characterizationsthrough scaling relations. Australian and New Zealand J.Statist. 49 (2), 115-135.I Navarro and Sarabia (2010). Alternative de�nitions ofbivariate equilibrium distributions. JSPI 140, 2046-2056.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Guillamon, Navarro and Ruiz (1998). Kernel densityestimation using weighted data. Comm. Statist. Theory andMethods 27, 2123-2135.I Navarro, Ruiz and del Aguila (2001). Parametric estimationfrom weighted samples. Biometrical J. 43, 297-311.I Navarro, Ruiz and del Aguila (2003). How to detect biasedsamples?. Biometrical J. 44, 742-763.I Pakes, Navarro, Ruiz and del Aguila (2003). Characterizationsusing weighted distributions. JSPI 116, 389-420.

I Navarro, Ruiz and del Aguila (2006). Multivariate weighteddistributions. Statist. 40 (1), 51-54.I Pakes and Navarro (2007). Distributional characterizationsthrough scaling relations. Australian and New Zealand J.Statist. 49 (2), 115-135.I Navarro and Sarabia (2010). Alternative de�nitions ofbivariate equilibrium distributions. JSPI 140, 2046-2056.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Guillamon, Navarro and Ruiz (1998). Kernel densityestimation using weighted data. Comm. Statist. Theory andMethods 27, 2123-2135.I Navarro, Ruiz and del Aguila (2001). Parametric estimationfrom weighted samples. Biometrical J. 43, 297-311.I Navarro, Ruiz and del Aguila (2003). How to detect biasedsamples?. Biometrical J. 44, 742-763.I Pakes, Navarro, Ruiz and del Aguila (2003). Characterizationsusing weighted distributions. JSPI 116, 389-420.I Navarro, Ruiz and del Aguila (2006). Multivariate weighteddistributions. Statist. 40 (1), 51-54.

I Pakes and Navarro (2007). Distributional characterizationsthrough scaling relations. Australian and New Zealand J.Statist. 49 (2), 115-135.I Navarro and Sarabia (2010). Alternative de�nitions ofbivariate equilibrium distributions. JSPI 140, 2046-2056.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Guillamon, Navarro and Ruiz (1998). Kernel densityestimation using weighted data. Comm. Statist. Theory andMethods 27, 2123-2135.I Navarro, Ruiz and del Aguila (2001). Parametric estimationfrom weighted samples. Biometrical J. 43, 297-311.I Navarro, Ruiz and del Aguila (2003). How to detect biasedsamples?. Biometrical J. 44, 742-763.I Pakes, Navarro, Ruiz and del Aguila (2003). Characterizationsusing weighted distributions. JSPI 116, 389-420.I Navarro, Ruiz and del Aguila (2006). Multivariate weighteddistributions. Statist. 40 (1), 51-54.I Pakes and Navarro (2007). Distributional characterizationsthrough scaling relations. Australian and New Zealand J.Statist. 49 (2), 115-135.

I Navarro and Sarabia (2010). Alternative de�nitions ofbivariate equilibrium distributions. JSPI 140, 2046-2056.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Guillamon, Navarro and Ruiz (1998). Kernel densityestimation using weighted data. Comm. Statist. Theory andMethods 27, 2123-2135.I Navarro, Ruiz and del Aguila (2001). Parametric estimationfrom weighted samples. Biometrical J. 43, 297-311.I Navarro, Ruiz and del Aguila (2003). How to detect biasedsamples?. Biometrical J. 44, 742-763.I Pakes, Navarro, Ruiz and del Aguila (2003). Characterizationsusing weighted distributions. JSPI 116, 389-420.I Navarro, Ruiz and del Aguila (2006). Multivariate weighteddistributions. Statist. 40 (1), 51-54.I Pakes and Navarro (2007). Distributional characterizationsthrough scaling relations. Australian and New Zealand J.Statist. 49 (2), 115-135.I Navarro and Sarabia (2010). Alternative de�nitions ofbivariate equilibrium distributions. JSPI 140, 2046-2056.Jorge Navarro; Biased samples (in honor of Prof. C.R. Rao)

Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Bayarri, M. J. and DeGroot, M. H. (1986). Information inselection models. Technical report.

I Arnold, B. C.; Castillo, E.; Sarabia, J. M. (2005).Distributions with conditionals in truncated weighted families.Statistics 39, no. 2, 133{147.I Crist�obal, J. A.; Alcal�a, J. T. (2000). Nonparametricregression estimators for length biased data. J. Statist. Plann.Inference 89 (2000), no. 1-2, 145{168.I Crist�obal, Jos�e A.; Alcal�a, Jos�e T. (2001). An overview ofnonparametric contributions to the problem of functionalestimation from biased data. Test 10, no. 2, 309{332.I Crist�obal, J. A.; Ojeda, J. L.; Alcal�a, J. T. (2004). Con�dencebands in nonparametric regression with length biased data.Ann. Inst. Statist. Math. 56, no. 3, 475{496.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Bayarri, M. J. and DeGroot, M. H. (1986). Information inselection models. Technical report.I Arnold, B. C.; Castillo, E.; Sarabia, J. M. (2005).Distributions with conditionals in truncated weighted families.Statistics 39, no. 2, 133{147.

I Crist�obal, J. A.; Alcal�a, J. T. (2000). Nonparametricregression estimators for length biased data. J. Statist. Plann.Inference 89 (2000), no. 1-2, 145{168.I Crist�obal, Jos�e A.; Alcal�a, Jos�e T. (2001). An overview ofnonparametric contributions to the problem of functionalestimation from biased data. Test 10, no. 2, 309{332.I Crist�obal, J. A.; Ojeda, J. L.; Alcal�a, J. T. (2004). Con�dencebands in nonparametric regression with length biased data.Ann. Inst. Statist. Math. 56, no. 3, 475{496.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Bayarri, M. J. and DeGroot, M. H. (1986). Information inselection models. Technical report.I Arnold, B. C.; Castillo, E.; Sarabia, J. M. (2005).Distributions with conditionals in truncated weighted families.Statistics 39, no. 2, 133{147.I Crist�obal, J. A.; Alcal�a, J. T. (2000). Nonparametricregression estimators for length biased data. J. Statist. Plann.Inference 89 (2000), no. 1-2, 145{168.

I Crist�obal, Jos�e A.; Alcal�a, Jos�e T. (2001). An overview ofnonparametric contributions to the problem of functionalestimation from biased data. Test 10, no. 2, 309{332.I Crist�obal, J. A.; Ojeda, J. L.; Alcal�a, J. T. (2004). Con�dencebands in nonparametric regression with length biased data.Ann. Inst. Statist. Math. 56, no. 3, 475{496.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Bayarri, M. J. and DeGroot, M. H. (1986). Information inselection models. Technical report.I Arnold, B. C.; Castillo, E.; Sarabia, J. M. (2005).Distributions with conditionals in truncated weighted families.Statistics 39, no. 2, 133{147.I Crist�obal, J. A.; Alcal�a, J. T. (2000). Nonparametricregression estimators for length biased data. J. Statist. Plann.Inference 89 (2000), no. 1-2, 145{168.I Crist�obal, Jos�e A.; Alcal�a, Jos�e T. (2001). An overview ofnonparametric contributions to the problem of functionalestimation from biased data. Test 10, no. 2, 309{332.

I Crist�obal, J. A.; Ojeda, J. L.; Alcal�a, J. T. (2004). Con�dencebands in nonparametric regression with length biased data.Ann. Inst. Statist. Math. 56, no. 3, 475{496.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Bayarri, M. J. and DeGroot, M. H. (1986). Information inselection models. Technical report.I Arnold, B. C.; Castillo, E.; Sarabia, J. M. (2005).Distributions with conditionals in truncated weighted families.Statistics 39, no. 2, 133{147.I Crist�obal, J. A.; Alcal�a, J. T. (2000). Nonparametricregression estimators for length biased data. J. Statist. Plann.Inference 89 (2000), no. 1-2, 145{168.I Crist�obal, Jos�e A.; Alcal�a, Jos�e T. (2001). An overview ofnonparametric contributions to the problem of functionalestimation from biased data. Test 10, no. 2, 309{332.I Crist�obal, J. A.; Ojeda, J. L.; Alcal�a, J. T. (2004). Con�dencebands in nonparametric regression with length biased data.Ann. Inst. Statist. Math. 56, no. 3, 475{496.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Marron, J. S.; de U~na-�Alvarez, J. (2004). SiZer for lengthbiased, censored density and hazard estimation. J. Statist.Plann. Inference 121, no. 1, 149{161.

I de U~na-�Alvarez, J.; Rodr��guez-Casal, A. (2007).Nonparametric estimation from length-biased data undercompeting risks. Comput. Statist. Data Anal. 51, no. 5,2653{2669.I de U~na-�Alvarez, J.; Saavedra, A. (2004). Bias and variance ofthe nonparametric MLE under length-biased censoredsampling: a simulation study. Comm. Statist. SimulationComput. 33, no. 2, 397{413I de U~na-�Alvarez, J.; Rodr��guez-Casal, A. (2006). Comparingnonparametric estimators for length-biased data. Comm.Statist. Theory Methods 35, no. 4-6, 905{919.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Marron, J. S.; de U~na-�Alvarez, J. (2004). SiZer for lengthbiased, censored density and hazard estimation. J. Statist.Plann. Inference 121, no. 1, 149{161.I de U~na-�Alvarez, J.; Rodr��guez-Casal, A. (2007).Nonparametric estimation from length-biased data undercompeting risks. Comput. Statist. Data Anal. 51, no. 5,2653{2669.

I de U~na-�Alvarez, J.; Saavedra, A. (2004). Bias and variance ofthe nonparametric MLE under length-biased censoredsampling: a simulation study. Comm. Statist. SimulationComput. 33, no. 2, 397{413I de U~na-�Alvarez, J.; Rodr��guez-Casal, A. (2006). Comparingnonparametric estimators for length-biased data. Comm.Statist. Theory Methods 35, no. 4-6, 905{919.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Marron, J. S.; de U~na-�Alvarez, J. (2004). SiZer for lengthbiased, censored density and hazard estimation. J. Statist.Plann. Inference 121, no. 1, 149{161.I de U~na-�Alvarez, J.; Rodr��guez-Casal, A. (2007).Nonparametric estimation from length-biased data undercompeting risks. Comput. Statist. Data Anal. 51, no. 5,2653{2669.I de U~na-�Alvarez, J.; Saavedra, A. (2004). Bias and variance ofthe nonparametric MLE under length-biased censoredsampling: a simulation study. Comm. Statist. SimulationComput. 33, no. 2, 397{413

I de U~na-�Alvarez, J.; Rodr��guez-Casal, A. (2006). Comparingnonparametric estimators for length-biased data. Comm.Statist. Theory Methods 35, no. 4-6, 905{919.


Biased samplesRenewal processesHow to detect biased samples?Appendix ConclusionsSome of my referencesOther referencesI Marron, J. S.; de U~na-�Alvarez, J. (2004). SiZer for lengthbiased, censored density and hazard estimation. J. Statist.Plann. Inference 121, no. 1, 149{161.I de U~na-�Alvarez, J.; Rodr��guez-Casal, A. (2007).Nonparametric estimation from length-biased data undercompeting risks. Comput. Statist. Data Anal. 51, no. 5,2653{2669.I de U~na-�Alvarez, J.; Saavedra, A. (2004). Bias and variance ofthe nonparametric MLE under length-biased censoredsampling: a simulation study. Comm. Statist. SimulationComput. 33, no. 2, 397{413I de U~na-�Alvarez, J.; Rodr��guez-Casal, A. (2006). Comparingnonparametric estimators for length-biased data. Comm.Statist. Theory Methods 35, no. 4-6, 905{919.


biased samples (in honor of prof. c.r. rao)...biased samples (in honor of prof. c.r. rao) jorge...

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